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Difference between revisions of "2025 AMC 8 Problems/Problem 25"
(→‎Solution 6 (Rigorous Representation Theoretical))
 
(66 intermediate revisions by 23 users not shown)
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−
==Problem==
+
== Problem ==
 +
 
 
Makayla finds all the possible ways to draw a path in a <math>5 \times 5</math> diamond-shaped grid. Each path starts at the bottom of the grid and ends at the top, always moving one unit northeast or northwest. She computes the area of the region between each path and the right side of the grid. Two examples are shown in the figures below. What is the sum of the areas determined by all possible paths?
 
Makayla finds all the possible ways to draw a path in a <math>5 \times 5</math> diamond-shaped grid. Each path starts at the bottom of the grid and ends at the top, always moving one unit northeast or northwest. She computes the area of the region between each path and the right side of the grid. Two examples are shown in the figures below. What is the sum of the areas determined by all possible paths?
 +
 +
<asy>
 +
// Asymptote by aoum
 +
unitsize(5mm);
 +
 +
path createpath(pair[][] P, int[] p) {
 +
  int i=0;
 +
  int j=0;
 +
  path dp = P[0][0];
 +
  for (int s=0; s<10; ++s) {
 +
    if (p[s] == 0) {++i;} else {++j;}
 +
    dp = dp--P[i][j];
 +
  }
 +
  return(dp);
 +
}
 +
 +
pair A[][], B[][];
 +
for (int i=0; i<6; ++i) {
 +
  A[i] = new pair[];
 +
  B[i] = new pair[];
 +
  for (int j=0; j<6; ++j) {
 +
    A[i].push(rotate(45)*(i,j));
 +
    B[i].push(shift(9,0)*A[i][j]);
 +
  }
 +
}
 +
 +
int[] p = {0,0,1,1,1,0,0,1,1,0};
 +
path pA = createpath(A,p);
 +
 +
int[] q = {1,0,0,0,1,1,1,1,0,0};
 +
path qB = createpath(B,q);
 +
 +
fill(pA--A[5][0]--cycle,lightgray);
 +
fill(qB--B[5][0]--cycle,lightgray);
 +
 +
for (int i=0; i<6; ++i) {
 +
  draw(A[i][0]--A[i][5],gray);
 +
  draw(B[i][0]--B[i][5],gray);
 +
  draw(A[0][i]--A[5][i],gray);
 +
  draw(B[0][i]--B[5][i],gray);
 +
}
 +
 +
draw(pA,black+2bp);
 +
draw(qB,black+2bp);
 +
 +
dot(A[0][0],black+5bp);
 +
dot(A[5][5],black+5bp);
 +
dot(B[0][0],black+5bp);
 +
dot(B[5][5],black+5bp);
 +
 +
label("$\mathrm{area} = 11$", A[0][0], S);
 +
label("$\mathrm{area} = 13$", B[0][0], S);
 +
</asy>
  
 
<math>\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 3150 \qquad \textbf{(C)}\ 3840 \qquad \textbf{(D)}\ 4730 \qquad \textbf{(E)}\ 5050</math>
 
<math>\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 3150 \qquad \textbf{(C)}\ 3840 \qquad \textbf{(D)}\ 4730 \qquad \textbf{(E)}\ 5050</math>
  
 
==Solution 1==
 
==Solution 1==
−
Consider a given path. The region below it, and hence its area, is accounted for in exactly one way (that is, with the given path; there are no other ways to get the same region below the path). However, the region ABOVE the path is also accounted for by exactly one path (the one found by rotating the entire grid <math>180^{\circ}</math>)! Hence, between these two paths, the entire area of the grid (that being <math>25</math> square units) is accounted for. The total number of paths is <math>\binom{10}{5}=252</math> (we choose <math>5</math> of our steps to be northeaast steps, determining the northwest steps). Each pair is accounted for exactly twice (one for each member of the pair). Hence, our answer is <cmath>252 \cdot \frac{1}{2} \cdot 25 = \boxed{\textbf{(B)}~3150}.</cmath>
+
 
 +
'''Step 1:''' To find the total number of paths, observe that all paths will have <math>10</math> total steps. We have to choose which <math>5</math> of these steps will be NE (the rest will be NW). So the total number of paths is <math>\binom{10}{5}</math>.  
 +
The formula for [[Combination|combinations]] is: <math>\binom{n}{r} = \frac{n!}{r!(n-r)!}</math> and <math>\binom{10}{5} = \frac{10!}{5!\times5!}=252</math>.
 +
 
 +
'''Step 2:''' Each path splits the total area of <math>25</math> in two parts. So, for any path that gives <math>area = A</math>, you can find a unique ‘sister’ path that has an <math>area = 25-A</math> (in other words, the pair of paths have a combined area of 25).  Possible ways to define the ‘sister’ path are:
 +
*Rotate the entire grid <math>180^{\circ}</math>
 +
*Swap each step of the original paths (for example, each NW becomes NE) (this is a reflection over the diagonal)
 +
 
 +
'''Step 3:''' There are a few ways to get from this observation to the total area:
 +
*There are <math>252/2 = 126</math> pairs of such paths, and the total area of each pair is <math>25</math>.  So the total area given by all paths is <math>126 \times 25</math>.  
 +
*Each of the <math>252</math> paths gives an area of <math>25</math> if you also count the ‘sister’ paths. Since each ‘sister’ path is also one of the <math>252</math>, you have to divide by <math>2</math> to avoid double counting.  So the total area given by all paths is <math>\frac{252 \times 25}{2}</math>.
 +
*Note that the average area of two ‘sister’ paths is <math>\frac{25}{2}</math>, so you can think about every path having this area ''on average''.  So the total area given by all paths is <math>252 \times \frac{25}{2}</math>.
 +
 
 +
The final answer is <math>\boxed{\textbf{(B)}~3150}.</math>
 +
 
 +
Note : This problem has a  bijection (or 1-1 correspondence) , check out [https://artofproblemsolving.com/store/book/intermediate-counting Intermediate Counting & Probability, Chapter 4], and [https://artofproblemsolving.com/store/book/intro-counting Introduction to Counting & Probability, Chapter 5]
 +
 
 +
 
 +
~ cxsmi<BR>
 +
~ aleyang<BR>
 +
~ MathCosine<BR>
 +
~ [[User:Aoum|aoum]]
 +
 
 +
==Solution 2==
 +
 
 +
If we test this problem on a smaller <math>2 \times 2</math> diamond, we have <math>6</math> ways to go from <math>A</math> to <math>B</math>, and the total area is <math>0 + 1 + 2 + 2 + 3 + 4 = 12</math>, so the average area is <math>\frac{12}{6} = 2</math>, which is also the area of the diamond <math>2 \times 2 = 4</math> divided by 2. If we assume this is true for a <math>5 \times 5</math> diamond, then the average area is <math>\frac{25}{2}</math>. The number of paths from <math>A</math> to <math>B</math> is <math>\binom{10}{5} = 252</math>, and <math>252 \cdot \frac{25}{2} = \boxed{\textbf{(B)}~3150}</math>.
 +
 
 +
~alwaysgonnagiveyouup
 +
 
 +
==Solution 3 Easier to Motivate==
 +
 
 +
Most other solutions don't explain how they got all the cases, as well as require an insight that's somewhat hard to think of, so I'll explain another in detail.
 +
 
 +
If we rotate the grid <math>45</math> degrees clockwise we can have a 5x5 grid where we can move up and right. There can only be one horizontal line segment in each column at a certain y coordinate. We'll denote the y coordinates as <math>0,1,2,3,4,5</math>. Once we choose the y coordinates for each column, we have a unique path. However, we can't move down which means that columns cannot have a higher y coordinate than the ones to the right. This is the same as distributing 5 balls in six boxes labeled <math>0,1,2,3,4,5</math>. For example, if we get 2 in 0, 2 in 1, and 1 in 4, then the order would be <math>0,0,1,1,4</math>. This is one unique path, and the total number of paths is represented by 6+5-1 choose 6-1 = 252. Also, the sum of the y-coordinates represents the area, which means we want the average sum of the y-coordinates. This is <math>5*(0+5)/2 = 25/2</math>, and <math>25/2 * 252 = \fbox{3150}</math>.
 +
 
 +
~Bread10
 +
 
 +
==Solution 4==
 +
 
 +
As found, there are <math>252</math> total paths. However, we can take advantage of symmetry here. We consider the total area of the left side of the grid in each of the <math>252</math> paths. Since the total area of the left side of the grid in each of the <math>252</math> paths is equal to the total area of the right side of the grid in each of the <math>252</math> paths, and the two total areas sum to <math>(252)(25) = 6300</math>, then the total area of the right side of the grid is <math>6300/2 = \fbox{3150}</math>.
 +
 
 +
==Solution 5 (q-analog)==
 +
 
 +
q-analog basic definitions are for <math>k\in\mathbb{N}</math>, then <cmath>[k]_q=1+q+\cdots+q^{k-1}</cmath> with <cmath>[0]_q=1</cmath> The q-factorial of <math>n\in\mathbb{N}</math> is defined as <cmath>[n]!_q=[n]_q\cdot[n-1]_q\cdots[0]_q</cmath> The q-binomial theorem states <cmath>\binom{n}{k}_q=\frac{[n]!_q}{[k]!_q[n-k]!_q}</cmath> and finally for q-analog of lattice paths aka q-lattice path, we have the q-lattice path theorem which states for any <math>a, b\in\mathbb{N}</math>, <cmath>\sum_{P\in L(a,b)}q^{\text{area}(P)}=\binom{a+b}{a}_q</cmath> To get the sum of the areas we have to differentiate the q-Lattice Path and evaluate at <math>q=1</math>. For this problem our lattice path is <math>L(5,5)</math> and we calculate the <math>q</math>-binomial as <math>q^{25}+q^{24}+2q^{23}+3q^{22}+5q^{21}+7q^{20}+9q^{19}+\cdots</math> and take the derivative <math>\left.\frac{d}{dq}\binom{10}{5}_q\right|_{q=1}</math> to get <math>25q^{24}+24q^{23}+46q^{22}+66q^{21}+105q^{20}+140q^{19}+171q^{18}+\cdots</math> and then plug in the 1 to finally get <math>\fbox{3150}</math>. ~Lopkiloinm
 +
 
 +
shoutout to https://enumeration.ca/
 +
 
 +
==Solution 6 (Rigorous Representation Theoretical)==
 +
 
 +
Let <math>\mathcal P_{a,b}</math> be the set of lattice paths from <math>(0,0)</math> to <math>(a,b)</math>
 +
using north steps <math>N=(0,1)</math> and east steps <math>E=(1,0)</math>.
 +
There are
 +
<cmath>
 +
|\mathcal P_{a,b}|=\binom{a+b}{a}
 +
</cmath>
 +
such paths.
 +
 
 +
For a path <math>P\in\mathcal P_{a,b}</math> list its steps as the word
 +
<math>\mathbf w=w_1\cdots w_{a+b}</math> with each <math>w_i\in\{E,N\}</math>.
 +
Define the lattice area
 +
<cmath>
 +
\operatorname{area}(P)=\sum_{i=1}^{a}y_i(P),
 +
</cmath>
 +
where <math>y_i(P)</math> is the <math>y</math>-coordinate just before the <math>i</math>-th <math>E</math>.
 +
Equivalently,
 +
<cmath>
 +
\operatorname{area}(\mathbf w)=
 +
\iota(\mathbf w):=
 +
\bigl|\{(i,j):i<j,\;w_i=E,\;w_j=N\}\bigr|.
 +
</cmath>
 +
 
 +
The symmetric group <math>S_{a+b}</math> acts by permuting positions of the word, and the action is transitive; in fact
 +
<cmath>
 +
\mathcal P_{a,b}\cong
 +
S_{a+b}\big/\!\bigl(S_a\times S_b\bigr),
 +
\qquad
 +
M:=\mathbb C[\mathcal P_{a,b}]\cong
 +
\operatorname{Ind}^{S_{a+b}}_{S_a\times S_b}\mathbf 1.
 +
</cmath>
 +
 
 +
Fix an ordered pair of positions <math>(i,j)</math> with <math>i<j</math>.
 +
Among all words, exactly half have <math>(w_i,w_j)=(E,N)</math> and half have <math>(N,E)</math>.
 +
Hence
 +
<cmath>
 +
\sum_{\mathbf w\in\mathcal P_{a,b}}\iota(\mathbf w)
 +
=\binom{a+b}{2}\,\binom{a+b-2}{a-1}
 +
=\frac{ab}{2}\,\binom{a+b}{a},
 +
</cmath>
 +
since for each ordered pair <math>(i,j)</math> we set <math>w_i=E</math>, <math>w_j=N</math>
 +
and distribute the remaining <math>a-1</math> <math>E</math>’s and <math>b-1</math> <math>N</math>’s among the
 +
other <math>a+b-2</math> positions in <math>\binom{a+b-2}{a-1}</math> ways.
 +
 
 +
Therefore
 +
<cmath>
 +
\boxed{\;
 +
\sum_{P\in\mathcal P_{a,b}}\operatorname{area}(P)
 +
      =\binom{a+b}{a}\,\frac{ab}{2}}
 +
</cmath>
 +
and the average lattice area under such paths is <math>ab/2</math>.
 +
 
 +
==Video Solution ==
 +
https://youtu.be/VP7g-s8akMY?si=VWOqMx55d5ctta1P&t=4042
 +
~hsnacademy
 +
 
 +
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=jTTcscvcQmI
 +
 
 +
==Video Solution by Thinking Feet==
 +
https://youtu.be/PKMpTS6b988
 +
 
 +
==Video Solution by Dr. David==
 +
https://youtu.be/sdZ4x5CBhIc
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
Problem 25 Only: https://youtu.be/ewyAsvVl2vY
 +
Whole Test: https://youtu.be/RYWx-JD7E0Y
 +
 
 +
~IceMatrix
 +
 
 +
== See Also ==
 +
 
 +
{{AMC8 box|year=2025|num-b=24|after=Last Problem}}
 +
{{MAA Notice}}
 +
 
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 21:58, 27 June 2025

Problem

Makayla finds all the possible ways to draw a path in a $5 \times 5$ diamond-shaped grid. Each path starts at the bottom of the grid and ends at the top, always moving one unit northeast or northwest. She computes the area of the region between each path and the right side of the grid. Two examples are shown in the figures below. What is the sum of the areas determined by all possible paths?

[asy] // Asymptote by aoum unitsize(5mm); path createpath(pair[][] P, int[] p) { int i=0; int j=0; path dp = P[0][0]; for (int s=0; s<10; ++s) { if (p[s] == 0) {++i;} else {++j;} dp = dp--P[i][j]; } return(dp); } pair A[][], B[][]; for (int i=0; i<6; ++i) { A[i] = new pair[]; B[i] = new pair[]; for (int j=0; j<6; ++j) { A[i].push(rotate(45)*(i,j)); B[i].push(shift(9,0)*A[i][j]); } } int[] p = {0,0,1,1,1,0,0,1,1,0}; path pA = createpath(A,p); int[] q = {1,0,0,0,1,1,1,1,0,0}; path qB = createpath(B,q); fill(pA--A[5][0]--cycle,lightgray); fill(qB--B[5][0]--cycle,lightgray); for (int i=0; i<6; ++i) { draw(A[i][0]--A[i][5],gray); draw(B[i][0]--B[i][5],gray); draw(A[0][i]--A[5][i],gray); draw(B[0][i]--B[5][i],gray); } draw(pA,black+2bp); draw(qB,black+2bp); dot(A[0][0],black+5bp); dot(A[5][5],black+5bp); dot(B[0][0],black+5bp); dot(B[5][5],black+5bp); label("$\mathrm{area} = 11$", A[0][0], S); label("$\mathrm{area} = 13$", B[0][0], S); [/asy]

$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 3150 \qquad \textbf{(C)}\ 3840 \qquad \textbf{(D)}\ 4730 \qquad \textbf{(E)}\ 5050$

Solution 1

Step 1: To find the total number of paths, observe that all paths will have $10$ total steps. We have to choose which $5$ of these steps will be NE (the rest will be NW). So the total number of paths is $\binom{10}{5}$. The formula for combinations is: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ and $\binom{10}{5} = \frac{10!}{5!\times5!}=252$.

Step 2: Each path splits the total area of $25$ in two parts. So, for any path that gives $area = A$, you can find a unique ‘sister’ path that has an $area = 25-A$ (in other words, the pair of paths have a combined area of 25). Possible ways to define the ‘sister’ path are:

  • Rotate the entire grid $180^{\circ}$
  • Swap each step of the original paths (for example, each NW becomes NE) (this is a reflection over the diagonal)

Step 3: There are a few ways to get from this observation to the total area:

  • There are $252/2 = 126$ pairs of such paths, and the total area of each pair is $25$. So the total area given by all paths is $126 \times 25$.
  • Each of the $252$ paths gives an area of $25$ if you also count the ‘sister’ paths. Since each ‘sister’ path is also one of the $252$, you have to divide by $2$ to avoid double counting. So the total area given by all paths is $\frac{252 \times 25}{2}$.
  • Note that the average area of two ‘sister’ paths is $\frac{25}{2}$, so you can think about every path having this area on average. So the total area given by all paths is $252 \times \frac{25}{2}$.

The final answer is $\boxed{\textbf{(B)}~3150}.$

Note : This problem has a bijection (or 1-1 correspondence) , check out Intermediate Counting & Probability, Chapter 4, and Introduction to Counting & Probability, Chapter 5


~ cxsmi
~ aleyang
~ MathCosine
~ aoum

Solution 2

If we test this problem on a smaller $2 \times 2$ diamond, we have $6$ ways to go from $A$ to $B$, and the total area is $0 + 1 + 2 + 2 + 3 + 4 = 12$, so the average area is $\frac{12}{6} = 2$, which is also the area of the diamond $2 \times 2 = 4$ divided by 2. If we assume this is true for a $5 \times 5$ diamond, then the average area is $\frac{25}{2}$. The number of paths from $A$ to $B$ is $\binom{10}{5} = 252$, and $252 \cdot \frac{25}{2} = \boxed{\textbf{(B)}~3150}$.

~alwaysgonnagiveyouup

Solution 3 Easier to Motivate

Most other solutions don't explain how they got all the cases, as well as require an insight that's somewhat hard to think of, so I'll explain another in detail.

If we rotate the grid $45$ degrees clockwise we can have a 5x5 grid where we can move up and right. There can only be one horizontal line segment in each column at a certain y coordinate. We'll denote the y coordinates as $0,1,2,3,4,5$. Once we choose the y coordinates for each column, we have a unique path. However, we can't move down which means that columns cannot have a higher y coordinate than the ones to the right. This is the same as distributing 5 balls in six boxes labeled $0,1,2,3,4,5$. For example, if we get 2 in 0, 2 in 1, and 1 in 4, then the order would be $0,0,1,1,4$. This is one unique path, and the total number of paths is represented by 6+5-1 choose 6-1 = 252. Also, the sum of the y-coordinates represents the area, which means we want the average sum of the y-coordinates. This is $5*(0+5)/2 = 25/2$, and $25/2 * 252 = \fbox{3150}$.

~Bread10

Solution 4

As found, there are $252$ total paths. However, we can take advantage of symmetry here. We consider the total area of the left side of the grid in each of the $252$ paths. Since the total area of the left side of the grid in each of the $252$ paths is equal to the total area of the right side of the grid in each of the $252$ paths, and the two total areas sum to $(252)(25) = 6300$, then the total area of the right side of the grid is $6300/2 = \fbox{3150}$.

Solution 5 (q-analog)

q-analog basic definitions are for $k\in\mathbb{N}$, then \[[k]_q=1+q+\cdots+q^{k-1}\] with \[[0]_q=1\] The q-factorial of $n\in\mathbb{N}$ is defined as \[[n]!_q=[n]_q\cdot[n-1]_q\cdots[0]_q\] The q-binomial theorem states \[\binom{n}{k}_q=\frac{[n]!_q}{[k]!_q[n-k]!_q}\] and finally for q-analog of lattice paths aka q-lattice path, we have the q-lattice path theorem which states for any $a, b\in\mathbb{N}$, \[\sum_{P\in L(a,b)}q^{\text{area}(P)}=\binom{a+b}{a}_q\] To get the sum of the areas we have to differentiate the q-Lattice Path and evaluate at $q=1$. For this problem our lattice path is $L(5,5)$ and we calculate the $q$-binomial as $q^{25}+q^{24}+2q^{23}+3q^{22}+5q^{21}+7q^{20}+9q^{19}+\cdots$ and take the derivative $\left.\frac{d}{dq}\binom{10}{5}_q\right|_{q=1}$ to get $25q^{24}+24q^{23}+46q^{22}+66q^{21}+105q^{20}+140q^{19}+171q^{18}+\cdots$ and then plug in the 1 to finally get $\fbox{3150}$. ~Lopkiloinm

shoutout to https://enumeration.ca/

Solution 6 (Rigorous Representation Theoretical)

Let $\mathcal P_{a,b}$ be the set of lattice paths from $(0,0)$ to $(a,b)$ using north steps $N=(0,1)$ and east steps $E=(1,0)$. There are \[|\mathcal P_{a,b}|=\binom{a+b}{a}\] such paths.

For a path $P\in\mathcal P_{a,b}$ list its steps as the word $\mathbf w=w_1\cdots w_{a+b}$ with each $w_i\in\{E,N\}$. Define the lattice area \[\operatorname{area}(P)=\sum_{i=1}^{a}y_i(P),\] where $y_i(P)$ is the $y$-coordinate just before the $i$-th $E$. Equivalently, \[\operatorname{area}(\mathbf w)= \iota(\mathbf w):= \bigl|\{(i,j):i<j,\;w_i=E,\;w_j=N\}\bigr|.\]

The symmetric group $S_{a+b}$ acts by permuting positions of the word, and the action is transitive; in fact \[\mathcal P_{a,b}\cong S_{a+b}\big/\!\bigl(S_a\times S_b\bigr), \qquad M:=\mathbb C[\mathcal P_{a,b}]\cong \operatorname{Ind}^{S_{a+b}}_{S_a\times S_b}\mathbf 1.\]

Fix an ordered pair of positions $(i,j)$ with $i<j$. Among all words, exactly half have $(w_i,w_j)=(E,N)$ and half have $(N,E)$. Hence \[\sum_{\mathbf w\in\mathcal P_{a,b}}\iota(\mathbf w) =\binom{a+b}{2}\,\binom{a+b-2}{a-1} =\frac{ab}{2}\,\binom{a+b}{a},\] since for each ordered pair $(i,j)$ we set $w_i=E$, $w_j=N$ and distribute the remaining $a-1$ $E$’s and $b-1$ $N$’s among the other $a+b-2$ positions in $\binom{a+b-2}{a-1}$ ways.

Therefore \[\boxed{\; \sum_{P\in\mathcal P_{a,b}}\operatorname{area}(P) =\binom{a+b}{a}\,\frac{ab}{2}}\] and the average lattice area under such paths is $ab/2$.

Video Solution

https://youtu.be/VP7g-s8akMY?si=VWOqMx55d5ctta1P&t=4042 ~hsnacademy

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution by Dr. David

https://youtu.be/sdZ4x5CBhIc

Video Solution by TheBeautyofMath

Problem 25 Only: https://youtu.be/ewyAsvVl2vY Whole Test: https://youtu.be/RYWx-JD7E0Y

~IceMatrix

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AMC_8_Problems/Problem_25&oldid=252552"